A Cam is to give following motion to a knife-edged follower.

i) Follower move to rise through 40 mm during 90° rotation of the Cam.

ii) Follower to dwell for next 45° rotation of the Cam.

iii) Follower to return to its original position during next 120° rotation.

iv) Follower to dwell for the remaining period.

The follower moves with simple harmonic motion during both rise and return stroke. The least radius of Cam is 50 mm. What will be the maximum velocity of the follower during rise if the angular velocity of the Cam is 30 rad/sec?

Concept:

For Simple Harmonic Motion (SHM), the displacement of a follower is given by:

\( s = \frac{h}{2} \left(1 - \cos\left(\frac{\pi \theta}{\beta}\right)\right) \)

Where,

• \( h = 40 \, \text{mm} \) (total rise),
• \( \beta = 90^\circ = \frac{\pi}{2} \, \text{rad} \) (angle for rise),
• \( \omega = 30 \, \text{rad/s} \) (cam angular velocity).

The velocity is the time derivative of displacement:

\( v = \frac{ds}{dt} = \frac{h}{2} \times \frac{\pi}{\beta} \times \omega \times \sin\left(\frac{\pi \theta}{\beta}\right) \)

The maximum velocity occurs when \( \sin\left(\frac{\pi \theta}{\beta}\right) = 1 \), thus:

\( v_{\text{max}} = \frac{h}{2} \times \frac{\pi}{\beta} \times \omega \)

Calculation:

Substitute the values:

\( v_{\text{max}} = \frac{40}{2} \times \frac{\pi}{\pi/2} \times 30 = 20 \times 2 \times 30 = 1200 \, \text{mm/s} \)

Convert to m/s:

\( v_{\text{max}} = \frac{1200}{1000} = \boxed{1.2 \, \text{m/s}} \)